If 2x − 3y = 7 and (a + b)x − (a + b − 3)y = 4a + b

The given system of equations are

$2 x-3 y=7$

$(a+b) x-(a+b-3) y=4 a+b$

For coincident lines , infinite number of solution

$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b 2}=\frac{c_{1}}{c_{2}}$

$\Rightarrow \frac{2}{(a+b)}=\frac{-3}{-(a+b-3)}=\frac{7}{4 a+b}$

$\Rightarrow 2(a+b-3)=3(a+b)$

$\Rightarrow 2 a+2 b-6=3 a+3 b$

$\Rightarrow 2 a+2 b-3 a-3 b=6$

$\Rightarrow-a-b=6$

$\Rightarrow a+b=-6---(i)$

$3(4 a+b)=7(a+b-3)$

$\Rightarrow 12 a+3 b=7 a+7 b-21$

$\Rightarrow 5 a-4 b=-21---(i i)$

multiply equation $(i)$ by 5, we get $5 a+5 b=-30---(i i i)$

$\operatorname{subtract}(i i)$ from $(i i i)$

$(5 a+5 b)-(5 a-4 b)=-30+21$

$\Rightarrow 5 a+5 b-5 a+4 b=-9$

$\Rightarrow 9 b=-9$

$\Rightarrow b=-1$

substitute $b=-1$ in equation $(1)$

$a+(-1)=-6$

$\Rightarrow a=-6+1=-5$

Option A:

$a+5 b=0$

$-5+5(-1)=-5-5=-10 \neq 0$

Option B:

$5 a+b=0$

$5(-5)+(-1)=-25-1=-26 \neq 0$

Option.C:

$a-b=0$

$-5-(-1)=-4 \neq 0$

None of the option satisfies the values.